# Homework Help: Electric Field To Potential

1. Mar 21, 2017

### Arman777

1. The problem statement, all variables and given/known data
I have a potential Value like $V=V(x,y,z)+C$
I found $\vec E$ using partial derivative, like $\vec E=((-∂V/∂x)i+(-∂V/∂y)j+(-∂V/∂z) k)$
Theres two position vectors,
$\vec r_{a}=2i$
$\vec r_{b}=j+k$
We need to find $V_{ba}=?$
2. Relevant equations
$V_b-V_a=-\int_{r_{a}}^{r_{b}} \vec E⋅d\vec r$
$V_r=V(x_i,y_i,z_i)$ where $r=(x_i,y_i,z_i)$

3. The attempt at a solution
Ok I found E but since we are taking partial derivative the constant term disappeared.

I can find from $V_b=V(0,1,1)$ and $V_a=V(2,0,0)$ and the difference will be $V_b-V_a=V_{ba}$

But If ı try to do this from $V_b-V_a=-\int_{r_{a}}^{r_{b}} \vec E⋅d\vec r$ using this.How can I approach the question.$\vec E$ is a function of $x,y,z$ but we need a function of $\vec r$

I mean the confusing part is,
$V_b-V_a=-\int_{r_{a}}^{r_{b}} ((-∂V/∂x)i+(-∂V/∂y)j+(-∂V/∂z) k)⋅d\vec r$

How can I take integral in this case ?

I ll do $\vec E⋅\vec r_a-\vec E⋅\vec r_b$ ??

And is my approach or answer is true..? , Is a constant term here makes a diffference ?

2. Mar 21, 2017

### BvU

Hi,

If you have an expression for V, why go the long way via $\vec E$ and integration if you can simply take $V_b-V_a$ ?

3. Mar 21, 2017

### Arman777

Just curiosity :)

4. Mar 21, 2017

### BvU

Write $d\vec r$ as $dx \, {\bf\hat\imath} + dy \, {\bf\hat\jmath}+ dz\, {\hat k}$ and write out the dot product to give you three terms of the integrand in three integrals...

5. Mar 21, 2017

### Arman777

so $V_b-V_a=-\int_{x=0}^{2}\int_{y=0}^{1}\int_{z=0}^{1} ((-∂V/∂x)+(-∂V/∂y)+(-∂V/∂z))dxdydz$

6. Mar 21, 2017

### Arman777

or maybe
$V_b-V_a=-\int_{x=0}^{2}((-∂V/∂x)dx+\int_{y=0}^{1}(-∂V/∂y)dy+\int_{z=0}^{1}(-∂V/∂z))dz$

7. Mar 21, 2017

### Arman777

Are these two integral same ?

8. Mar 21, 2017

### BvU

Try an example, e.g. $V(x, y, z) = 2x+3y+4z + 1199$

9. Mar 21, 2017

### BvU

No. dxdydz is a volume integral. That's not the idea...

10. Mar 21, 2017

### Arman777

I see you are right
But it must be the
I did your example and it gave me the same thing.I understand the idea.But the constant dissapered which thats bothers.Or maybe it didnt.

And thanks

11. Mar 21, 2017

### Arman777

I think in the calculating V difference, the constant has no affect.Of course it doesnt thats the logical thing...

12. Mar 21, 2017

### Arman777

Ok thanks a lot again

13. Mar 21, 2017

### BvU

Correct: a potential is basically unnoticeable. Only potential differences bring about something that can be sensed.